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# Prove!

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If V={(x,y)|x,y ∈ R} with acts (x1,y1)@(x2,y2)=$$(\sqrt[3]{(x1^3+x2^3)},\sqrt[3]{(y1^3+y2^3)})$$ and a\$(x1,y1)=(ax1,ay1) , a ∈ R.

Prove V with this acts is NOT vector space.Thank you!

Nov 12, 2018

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are you sure it's not?

it's addition has commutativity and associativity

(0,0) is the clear additive identity and (-x,-y) is the additive inverse of (x,y)

the scalar properties are identical to the Euclidean vector space

I'm not seeing why this isn't a vector space.

remember the cube and cube root form a bijection

Nov 12, 2018
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This is not a vector space. Exactly one of the axioms of a vector space is not satisfied.

Guest Nov 12, 2018
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Rom yes im sure the exercise say it's not.For this reason i can't solve the exercise

Dimitristhym  Nov 12, 2018
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Guest "Exactly one of the axioms of a vector space is not satisfied"

Witch is this axiom?

Dimitristhym  Nov 12, 2018
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Ok.  Distribution of scalar mutliplication with respect to field addition doesn't hold.

we should have

$$(a+b)\vec{v} = a \vec{v} \oplus b \vec{v}$$

let's take a look

$$(a+b)\vec{v} =\{(a+b)v_x,(a+b)v_y\}\\ a\vec{v} \oplus b \vec{v} =\left \{\sqrt[3]{a^3 v_x^3+b^3 v_x^3}, \sqrt[3]{a^3 v_y^3+b^3 v_y^3 }\right \}=\\ \left\{\sqrt[3]{a^3+b^3}v_x,~\sqrt[3]{a^3+b^3}v_y \right\}\\ \sqrt[3]{a^3+b^3} \neq a+b$$

Rom  Nov 13, 2018
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Thank you very much Rom!

Dimitristhym  Nov 13, 2018